Set $n=2^k$ (for some integer $k$) and let $D={\rm diag}(d_1,d_2,\cdots,d_n)$ and $D' = {\rm diag}(d_1', d_2 ,\cdots, d_n')$ be two diagonal matrices in $\mathbb C^{n \times n}$. Let us also presume all diagonal entries are positive: $d_i >0$ and $d_i'>0$. I am curious about the $n \times n$ matrix
$$ A = \frac{1}{n} \; D H_n D' H_n $$
where $H_n$ is the Sylvester-Hadamard matrix of order $n=2^k$ (also sometimes called a Walsh matrix). Explicitly:
$$ H_{2^k} = \left( \begin{array}{cc} H_{2^{k-1}} & H_{2^{k-1}} \\ H_{2^{k-1}} & -H_{2^{k-1}} \end{array} \right) = H_2 \otimes H_{2^{k-1}} \qquad \textrm{where } H_2 = \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right).$$
Question: Is there any hope for analytically determining the eigenvalues of $A$?
Some comments about the structure of $A$:
Note that $A$ is diagonalizable, since by a similarity transform it is equivalent to the symmetric real matrix $\frac{1}{n} \sqrt{D} H D' H \sqrt{D}$. Moreover, since the inverse of this matrix is simply $\frac{1}{n} \sqrt{D}^{-1} H {(D')}^{-1} H \sqrt{D}^{-1}$, we learn that if the eigenvalues of $A$ are denoted as $\{\lambda_i(d_1,d_2,\cdots,d_n;d_1',d_2',\cdots,d_n')\}_{i =1,2,\cdots, n}$, then
$$\lambda_i(d_1^{-1},d_2^{-1},\cdots,d_n^{-1};(d_1')^{-1},(d_2')^{-1},\cdots,(d_n')^{-1}) = \left( \lambda_i(d_1,d_2,\cdots,d_n;d_1',d_2',\cdots,d_n') \right)^{-1}.$$
We can also see that $\lambda_i(\{ d_j \}; \{d_j'\}) = \lambda_i(\{ d_j' \}; \{d_j\})$.
The form of $A$ also suggests other relations, such as
$$ {\rm tr}(A) = \frac{1}{n} {\rm tr}(D) {\rm tr}(D') = \frac{1}{n} \sum_{i,j=1}^n d_i d_j' $$
and
$$ {\rm det}(A) = {\rm det}(D) {\rm det}(D') = \prod_{i,j=1}^n d_i d_j'. $$
All this is still a far-cry from determining the spectrum of $A$, but I am wondering if the form of $A$ is special enough to hope that the spectrum of $A$ is tractable, or whether there is a good reason to suspect it is not.
